Question

\[ \text{The conductivity of }0.001M\text{ solution of acetic acid is }3.905\times10^{-5}\;S\,cm^{-1}.\text{ Calculate its molar conductivity and degree of dissociation }(\alpha). \]
\[ \lambda^\circ_{CH_3COO^-}=40.9\;S\,cm^2mol^{-1} \]
\[ \lambda^\circ_{H^+}=349.6\;S\,cm^2mol^{-1} \]

Hint:

For weak electrolytes, \(\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}\). Salt bridge prevents charge build-up in half-cells.

Answer 1

Step 1: Write the given data.
Conductivity: $\kappa = 3.905 \times 10^{-5}$ S cm$^{-1}$. Concentration: $C = 0.001$ M. $\lambda^\circ_{H^+} = 349.6$ S cm$^2$ mol$^{-1}$. $\lambda^\circ_{CH_3COO^-} = 40.9$ S cm$^2$ mol$^{-1}$.
Step 2: Calculate molar conductivity $\Lambda_m$.
\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{3.905 \times 10^{-5} \times 1000}{0.001} = \frac{3.905 \times 10^{-2}}{10^{-3}} = 39.05 \text{ S cm}^2 \text{ mol}^{-1} \]
Step 3: Calculate $\Lambda_m^\circ$ using Kohlrausch's law.
For acetic acid ($CH_3COOH \rightarrow H^+ + CH_3COO^-$): \[ \Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} = 349.6 + 40.9 = 390.5 \text{ S cm}^2 \text{ mol}^{-1} \]
Step 4: Calculate degree of dissociation $\alpha$.
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{39.05}{390.5} = 0.10 \] So $\alpha = 0.10$ (10% dissociation of acetic acid at $0.001$ M).
Step 5: Interpret the result.
Acetic acid is a weak electrolyte. Only 10% of $CH_3COOH$ molecules are ionised at this concentration. As the solution is further diluted, $\alpha$ increases (Ostwald's dilution law). At infinite dilution, $\alpha \rightarrow 1$ (complete dissociation), and $\Lambda_m \rightarrow \Lambda_m^\circ$.
Step 6: State the final answers.
Molar conductivity: $\Lambda_m = 39.05$ S cm$^2$ mol$^{-1}$. Degree of dissociation: $\alpha = 0.10$ (10%). \[ \boxed{\Lambda_m = 39.05 \text{ S cm}^2 \text{ mol}^{-1};\; \alpha = 0.10} \]